Quadratics - Nautical miles and knots question

Question

Ship A is 50 nautical miles west of Ship B. Ship A is heading east at 10 knots and ship B is heading south at 5 knots. Find the minimum distance between the ships, and at what time it occurred

What are the steps to this question, and how would the solution look as a parabola?

Answer 1

HINT

1. Let's place the origin at the current location of $A$. Can you write down $A(t) = (x_a(t), y_a(t))$, the location of $A$ at time $t \ge 0$?
2. Now write down $B(t) = (x_b(t), y_b(t))$, the location of $B$ at time $t\ge 0$.
3. Use the distance formula to find the square distance between them at time $t$, say $D(t)$.
4. Minimize $D(t)$ over $t \ge 0$.

Answer 2

Since the two ships are traveling in directions at right angles to one another and we want to find the distance between them at any moment, it is convenient to choose the coordinate axes as the lines the ships are moving along. There is nothing in setting up the problem that will be complicated by taking the direction each ship is moving to be the "positive direction", so we may call due East the positive$-x \ $ direction and due South, the positive $-y \ $ direction.

We can also choose to make the origin the starting point of ship B, so that its subsequent position at $ \ t \ $ hours after the "initial moment" $ \ (t=0) \ $ in the problem is $ \ x_B \ = \ 0 \ , \ y_B \ = \ 5·t \ \ , $ since a knot is a speed of one nautical mile per hour. Ship A is then starting $ \ 50 \ $ nautical miles West of the origin and traveling Eastward, so its position at $ \ t \ $ hours is $ \ x_A \ = \ -50 + 10·t \ , \ y_A \ = \ 0 \ \ . $

We eventually wish to find out when the two ships are closest to one another and what that minimum separation is. Since separation $ \ s \ $ is never negative, we can use the square of the separation between the ships, rather than using the "raw" distance formula which requires taking a square-root; this is because $ \ s^2 \ $ will be smallest when $ \ s \ $ itself is smallest. The distance formula comes from the "Pythagorean theorem", so we write $$ s^2 \ \ = \ \ (x_B - x_A)^2 \ + \ (y_B - y_A)^2 \ \ = \ \ (0 - [-50 + 10·t])^2 \ + \ (5·t - 0)^2 $$ $$ = \ \ (-50 + 10·t)^2 \ + \ (5·t)^2 \ \ \text{knots}^2 \ \ .$$

This description of the set-up for the solution is in aid of answering your last question. When we multiply out the terms in the separation-squared equation, we obtain $$ s^2 \ \ = \ \ (100·t^2 \ - \ 1000·t \ + \ 2500) \ + \ (25·t^2) \ \ \text{knots}^2 \ \ . $$ Since both the polynomial for the square of the difference of the $ \ x-$coordinates and that for the square of the difference of the $ \ y-$coordinates are quadratic , their sum will also be quadratic (because the leading terms do not cancel).

The graph of the separation-squared as a function of time (in hours), $ s^2 \ = \ 125·t^2 \ - \ 1000·t \ + \ 2500 \ \ , $ is then the parabolic curve shown below. We see that the ships will be closest to each other at $ \ t = 4 \ $ hours, when $ \ s^2 \ = \ 500 \ \text{knots}^2 \ \ . $ We can also determine this by completing-the-square:

$$ s^2 \ \ = \ \ 125·(t^2 \ - \ 8·t) \ \ + \ 2500 \ \ = \ \ 125·(t^2 \ - \ 8·t \ + \ 16) \ \ + \ 2500 - 125·16 $$ $$ = \ \ 125·(t \ - \ 4)^2 \ \ + \ 500 \ \ . $$

The minimum distance between the ships is thus $ \ s_{min} \ = \ \sqrt{500} \ \approx \ 22.4 \ $ nautical miles.